Question

1-Given the function: \(y=\frac{x^{2}-3 x-4}{x^{2}-5 x+4}\), decide if \(f(x)=y\) is continuous or has a removable discontinuity, and find horizontal tond vertical asymptotes.

2 A-Use the definition \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) to prove that derivative of \(f(x)=\sqrt{4-x}\) is \(\frac{-1}{2 \sqrt{4-x}}\)

2 B- Evaluate the limit \(\lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\) for the given value of \(x\) and function \(f(x) .\)

$$ f(x)=\sin x, \quad x=\frac{\pi}{4} $$

3-Given the function: \(y=(x+4)^{3}(x-2)^{2}\), find y' and classify critical numbers very carefully using first derivative tess oniy.

4-For the function \(f(x)=\sin x+\sqrt{3} \cos x\), find absolute maximum, minimum, increasing, decreasing, concave up, and concave down over \([0,2 \pi]\).

5- Differentiate and simplify your answer

$$ y x=\sqrt{1+x y} $$

\(x \sin y+y \cos x=1\)

Find the \(\frac{d y}{d x}\) Using implicit differcntiation where \(\quad x+\tan (x+y)=x^{2} y^{3}\)

$$ y=\ln (\sec x+\tan x) $$

$$ \begin{aligned} &y=\sin ^{5}\left(1-x^{2}\right) \\ &y=\ln \left(\frac{\left(x^{2}-3\right)^{3}(x+4)^{4}}{(1-x)^{2}}\right) \end{aligned} $$

6-Use implicit differentiation to find \(\frac{dy}{d x}\) were \(y^{2}+x y+4 y=x^{3}+5\)

7-Differentiate: \(y=\left(\frac{x-1}{x+1}\right)^{5}\) simplify your answer

8- Find all the points on the graph of the function \(f(x)=2 \sin x+\sin ^{2} x\) at which the tangent line is horizontal, \(0 \leq \mathrm{x} \leq 2 \mathrm{~T}\).

Answer 1

1. Given function
y = (x^2 - 3x - 4)/(x^2 - 5x + 4)
y = (x - 4)(x + 1)/(x - 4)(x - 1)
now,
we can see, at x = 4, lim x -> 4 y = 5/3 , this is removable discontinuity
at x = 1, we see that
lim x-> 1- y = -inf and lim x-> 1+ y = +inf , hence this is an infinite discontinuity
the vertical asymptotes are when dy/dx = inf
dy/dx = [(x^2 - 5x + 4)(2x - 3) - (x^2 - 3x - 4)(2x - 5)]/(x^2 - 5x + 4)
so, this happens at x = 1,

for horizontal asymptote
lim x-> inf y = 1
lim x -> -inf y = 1
hence y = 1 is horizontal asymptote

2.A. df(x)/dx = lim h -> 0 ((f(x + h) - f(x))/h)
f(x) = sqrt(4 - x)
df(x)/dx = lim h -> 0 (sqrt(4 - x - h) - sqrt(4 - x))/h
df(x)/dx = lim h -> 0 (4 - x - h -4 + x)/h*(sqrt(4 - x - h) + sqrt(4 - x))
df(x)/dx = lim h -> 0 1/(sqrt(4 - x - h) + sqrt(4 - x))
df(x)/dx = -1/2(sqrt(4 - x))

2.B. f(x) = sin(x)
f'(x) = cos(x)
at x = pi/4, f'(pi/4) = cos(pi/4) = sin(pi/4) = 1/sqrt(2)

3. y = (x + 4)^3(x - 2)^2
y' = 3(x + 4)^2(x - 2)^2 + 2(x + 4)^3(x - 2)
so, critical points
y = 0
hence, x = -4, or x = 2
so (-4,0), (2,0) are critical points

and y' = 0
3(x + 4)^2(x - 2)^2 + 2(x + 4)^3(x - 2) = 0
3(x - 2) + 2(x + 4) = 0
3x - 6 + 2x + 8 = 5x + 2 = 0
x = -2/5 = -0.4
(-0.4, 268.73856) , x = -4, x = 2 are critical points as well
(-4, 0) , (2, 0)
hence,

maxima at x = -0.4,
minima at x = -4 and x = 2
value 0 at x = -4 and x = 2

4. f(x) = sin(x) + sqrt(3)*cos(x)   
f(x) = 2[sin(x)*sin(30) + cos(x)cos(30)]
f(x) = 2[cos(x - 30)]
  
so for [0, 2*pi] interval
for [0, pi/6] the function is increasing,
for [pi/6, 7pi/6], the function is decreasing,
from [7pi/6, pi] the function is increasing

absolute maximum is 2, amsolute minimum is -2
between [0, 2*pi/3] the function is concave down
between [2*pi/3 , 5*pi/3] the function is concave down
then between [5*pi/3, 2*pi] the function is concave up